Questa sfida riguarda la costruzione di una scacchiera in cui la dimensione quadrata, anziché essere costante su tutta la linea, segue una certa sequenza non decrescente, come descritto di seguito.

La scheda è definita in modo iterativo. Una tavola di dimensioni $n \times n$ viene ingrandita a dimensioni estendendola verso il basso ea destra di uno "strato" di quadrati di dimensione , dove è il massimo divisore di non superiore a . I quadrati nella diagonale sono sempre dello stesso colore.$(n+k)\times(n+k)$$k$$k$$n$$\sqrt{n}$

In particolare, considera la lavagna con i colori rappresentati come #e +.

1. Inizializza la scacchiera su

   #
   

2. La scheda finora ha dimensioni $1\times 1$ . L'unico divisore di $1$ è $1$ e non supera . Quindi prendiamo ed estendiamo la tavola aggiungendo uno strato di quadrati di dimensione , con in diagonale:$\sqrt{1}$$k=1$$1$#

   #+
   +#
   

3. La scheda costruita finora ha dimensioni . I divisori di sono e il divisore massimo non superiore a è . Quindi di nuovo , e la scheda è estesa a$2 \times 2$$2$$1,2$$\sqrt{2}$$1$$k=1$

   #+#
   +#+
   #+#
   

4. La dimensione è . . Esteso a$3 \times 3$$k=1$

   #+#+
   +#+#
   #+#+
   +#+#
   

5. La dimensione è . Ora k = 2 , perché 2 è il divisore massimo di 4 non superiore a √$4 \times 4$$k=2$$2$$4$$\sqrt 4$ . Si estende con uno strato di spessore$2$, formato da quadrati di dimensioni$2\times 2$, con colore#nella diagonale:

   #+#+##
   +#+###
   #+#+++
   +#+#++
   ##++##
   ##++##
   

6. Le dimensioni sono $6 \times 6$ . Ora $k=2$ . Estendere alle dimensioni $8 \times 8$ . Ora $k=2$ . Estendere alle dimensioni $10 \times 10$ . Ora $k=2$ . Estendere alle dimensioni $12 \times 12$ . Ora $k=3$ . Estendi alla taglia $15$ :

   #+#+##++##++###
   +#+###++##++###
   #+#+++##++#####
   +#+#++##++##+++
   ##++##++##+++++
   ##++##++##+++++
   ++##++##++#####
   ++##++##++#####
   ##++##++##++###
   ##++##++##+++++
   ++##++##++##+++
   ++##++##++##+++
   ###+++###+++###
   ###+++###+++###
   ###+++###+++###
   

Si noti come i quadrati aggiunti più di recente, delle dimensioni $3 \times 3$ , abbiano lati che coincidono parzialmente con quelli dei quadrati aggiunti in precedenza delle dimensioni $2 \times 2$ .

La sequenza formata dai valori di $k$ non è decrescente:

1 1 1 2 2 2 2 3 3 3 3 4 4 4 6 6 6 6 6 6 ...

e non sembra essere nell'OEIS. Tuttavia, la sua versione cumulativa, che è la sequenza di dimensioni della scheda, è A139542 (grazie a @Arnauld per averlo notato).

La sfida

Input : un numero intero positivo $S$ rappresenta il numero di strati nella scheda. Se preferisci, puoi anche ottenere $S-1$ invece di $S$ come input ( $0$ -indexed); vedi sotto.

Output : una rappresentazione in arte ASCII di una scheda con livelli $S$

• L'output può essere tramite STDOUT o un argomento restituito da una funzione. In questo caso può essere una stringa con newline, una matrice di caratteri 2D o una matrice di stringhe.

• Puoi costantemente scegliere due personaggi qualsiasi per rappresentare il tabellone.

• Puoi scegliere costantemente la direzione della crescita. Cioè, invece delle rappresentazioni di cui sopra (che crescono verso il basso e verso destra), è possibile produrre una qualsiasi delle sue versioni riflesse o ruotate.

• È consentito lo spazio trailing o iniziale (se l'output è attraverso STDOUT), purché lo spazio non sia uno dei due caratteri utilizzati per la scheda.

• Opzionalmente puoi usare l' input " $0$ -indexed "; cioè, prendi come input $S-1$ , che specifica una scheda con livelli $S$

Vince il codice più breve in byte.

Casi test

1:

#

3:

#+#
+#+
#+#

5:

#+#+##
+#+###
#+#+++
+#+#++
##++##
##++##

6:

#+#+##++
+#+###++
#+#+++##
+#+#++##
##++##++
##++##++
++##++##
++##++##

10:

#+#+##++##++###+++
+#+###++##++###+++
#+#+++##++#####+++
+#+#++##++##+++###
##++##++##+++++###
##++##++##+++++###
++##++##++#####+++
++##++##++#####+++
##++##++##++###+++
##++##++##+++++###
++##++##++##+++###
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###+++###+++###+++
###+++###+++###+++
###+++###+++###+++
+++###+++###+++###
+++###+++###+++###
+++###+++###+++###

15:

#+#+##++##++###+++###+++####++++####
+#+###++##++###+++###+++####++++####
#+#+++##++#####+++###+++####++++####
+#+#++##++##+++###+++#######++++####
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25:

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Gelatina , 40 31 byte

1SÆD>Ðḟ½ƊṀṭƲ³¡Äż$Ḷ:Ḃ^þ`ʋ/€ḷ""/Y

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$S-1$

Senza il finale Y, questo restituisce un elenco di elenchi di numeri interi, ma questo è fuori specifica per questa sfida.

Spiegazione

Questo programma funziona in tre fasi.

1. $k$$k$
2. Genera una scacchiera per ognuno di questi con la dimensione del riquadro di $k$
3. Lavora attraverso l'elenco delle scacchiere, sostituendo ogni volta la sezione in alto a sinistra della scacchiera successiva con quella esistente.

Fase 1

1                 | Start with 1
           Ʋ³¡    | Loop through the following the number of times indicated by the first argument to the program; this generates a list of values of k
 S                | - Sum
        Ɗ         | - Following three links as a monad 
  ÆD              |   - List of divisors
    >Ðḟ½          |   - Exclude those greater than the square root
         Ṁ        |   - Maximum
          ṭ       | - Concatenate this to the end of the current list of values of k 
              Äż$ | Zip the cumulative sum of the values of k with the values

Fase 2

      ʋ/€ | For each pair of k and cumulative sum, call the following as a dyad with the cumulative sum of k as the left argument and k as the right (e.g. 15, 3)
Ḷ         | - Lowered range [0, 1 ... , 13, 14]
 :        | - Integer division by k [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
  Ḃ       | - Mod 2 [0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0]
   ^þ`    | - Outer product using xor function and same argument to both side

Fase 3

   /  | Reduce using the following:
ḷ""   | - Replace the top left portion of the next matrix with the current one
    Y | Finally join by newlines

Tela , 34 32 byte

０#０⁸［#+¶+#ｘｘ＊ｙｘ＋ｍ⤢αｍ；ｎｌｗ√｛ｙ；％‽²Ｘ

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Python 2 , 217 215 212 byte

def f(x):
 b=['1'];n=1
 for i in range(x):P=max(j*(n%j<(j<=n**.5))for j in range(1,1+n));n+=P;b=[l+P*`j/P%2^i%2`for j,l in enumerate(b)];s=len(b[0]);b+=[((v*P+`1^int(v)`*P)*s)[:s]for v in b[0][len(b):]]
 return b

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0 indicizzato, usa 0e 1come caratteri

Python 2 , 184 178 176 169 169 byte

def h(j,a=['1'],R=range):
 for i in R(j):L=len(a);k=max(x for x in R(1,L+1)if(x*x<=L)>L%x);a=[a[m]+k*`(i+m/k)%2`for m in R(L)]+[((`i%2`*k+`~i%2`*k)*L)[:L+k]]*k
 return a

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Usi 1, 0per #, -; usa 0-indexing.

JavaScript (ES7), 164 byte

$0$#$1$+

n=>(b=[1],g=(a,w,d=w**.5|0)=>b[n]?a:w%d?g(a,w,d-1):g(a.concat(Array(d).fill(b.push(d)&&i++)),w+d))([0],i=1).map((_,y,a)=>a.map((_,x)=>(x/b[v=a[x>y?x:y]]^y/b[v])&1))

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Carbone , 37 byte

ＦＮ«≔⊕⌈Φ₂⊕Ｌυ¬﹪Ｌυ⊕κηＦη«ＰL⭆⊞Ｏυω§#+÷⁻κμη↙

Provalo online! Il collegamento è alla versione dettagliata del codice. 1-indicizzati. L'output cresce verso il basso e verso sinistra (in basso e a destra costa un byte aggiuntivo, ma può crescere per lo stesso conteggio byte). Spiegazione:

ＦＮ«

$S$

≔⊕⌈Φ₂⊕Ｌυ¬﹪Ｌυ⊕κη

$k\leq\sqrt{n+1}$$n=0$$k=1$

Ｆη«

$k$

ＰL⭆⊞Ｏυω§#+÷⁻κμη

#+#⊞Ｏυω$n$

↙

Sposta in basso e lasciati pronti per la riga successiva.

05AB1E , 43 42 byte

$G©ÐX‚ˆÑʒ®>t‹}àDU+}¯εÝ`θ÷ɨDδ^}RζεðKζðδK€θ

Ispirato @NickKennedy 'risposta gelatina s , e la porzione posteriore ζεðKζðδK€θè una porta da @Emigna ' 05AB1E risposta s qui .

Restituisce una matrice di 0anziché #e 1invece di+ .

$[2,n]$J,--no-lazy bandiera si deve stampare piuttosto la matrice risultante).

Spiegazione:

$                # Push 1 and the input
 G               # Loop the input - 1 amount of times:
  ©              #  Store the top of the stack in variable `r` (without popping)
   Ð             #  And triplicate the top as well
    X‚           #  Pair it with variable `X` (which is 1 by default)
      ˆ          #  And pop and store this pair in the global array
    Ñ            #  Get the divisors of the integer we triplicated
     ʒ    }à     #  Get the highest divisor which is truthy for:
         ‹       #   Where the divisor integer is smaller than
      ®>t        #   the square root of `r+1`
            DU   #  Store a copy of this largest filtered divisor as new variable `X`
              +  #  And add it to the triplicated integer
 }¯              # After the loop: push the global array
   ε             # Map each pair to:
    Ý θ          #  Convert the first value in the pair to a list in the range [0,n]
     `           #  and push both this list and the second value to the stack
       ÷         # Integer-divide each value in the list by the second value
        É        # Check for each value if it's even (1 if even; 0 if odd)
         ¨       # Remove the last item
          Dδ     # Loop double vectorized over this list:
            ^    #  And XOR the values with each other
   }R            # After the map: reverse the list of digit-matrices
     ζ           # Zip/transpose; swapping rows and columns, with a space as filler
      ε          # map each matrix to:
       ðK        #  Remove all spaces from the current matrix
         ζ       #  Zip/transpose with a space as filler again
          ðδK    #  Deep remove all spaces
             €θ  #  Then only leave the last values of each row
                 # (after which the resulting matrix of 0s and 1s is output implicitly)

Haskell, 149 146 byte

(iterate g["#"]!!)
g b|let e=(<$[1..d]);l=length b;d=last[i|i<-[1..l],i*i<=l,mod l i<1];m="+#"++m=(e$take(l+d)$e=<<'#':m)++zipWith(++)(e=<<e<$>m)b

Questo è 0 indicizzato, restituisce un elenco di stringhe e cresce verso l'alto e verso sinistra.

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(iterate g["#"]!!)                    -- start with ["#"], repeatedly add a layer
                                      -- (via function 'g'), collect all results in
                                      -- a list and index it with the input number

g b | let                             -- add a single layer to chessboard 'b'

 l=length b                           -- let 'l' be the size of 'b'
 d=last[i|i<-[1..l],i*i<=l,mod l i<1] -- let 'd' be the size of the new layer
 e=(<$[1..d])                         -- let 'e' be a functions that makes 'd'
                                      --   copies of it's argument
 m="#+"++m                            -- let 'm' be an infinite string of "+#+#+..."

 =                                    -- return
              zipWith(++)             --   concatenate pairwise
                         (e=<<e<$>m)  --   a list of squares made by expanding each
                                      --   char in 'm' to size 'd'-by-'d'
                                    b --   and 'b' (zipWith truncates the infinite
                                      --   list of squares to the length of 'b')
                                      --
           ++                         --   and prepend
                                      --
(e$take(l+d)$e=<<'#':m)               --   the top layer, i.e. a list of 'd' strings
                                      --   each with the pattern 'd' times '#'
                                      --   followed by 'd' times '+', etc., each
                                      --   shortened to the correct size of 'l'+'g'

Perl 6 , 156 144 155 154 byte

+11 per correggere un bug segnalato da nimi.

{$!=-1;join "
",(1,{my \k=max grep $_%%*,1.. .sqrt;++$!;flat .kv.map(->\i,\l {l~($!+i/k)%2+|0 x k}),substr(($!%2 x k~1-$!%2 x k)x$_,0,$_+k)xx k}...*)[$_]}

Circa basato sulla soluzione Python di Chas Brown . Prende S in modo zero indicizzato. Uscite 0e 1.

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